--- title: "Digital LPF to HPF transformation" subject: "Digital Signal Analysis and Processing" chapter: 6 key: "semester_7/digital_signal_analysis_and_processing:lowpass-to-highpass-transformation" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lowpass-to-highpass-transformation site: BCT Engineering --- # Digital LPF to HPF transformation Digital Signal Analysis and Processing · Chapter 6: IIR Filter Design - Asked in: 2070 Chaitra, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lowpass-to-highpass-transformation - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-6 ## Question A digital LPF with cut off frequency ωc = 0.2575π is given as H(z) = (0.1 + 0.4z⁻¹)/(1 − 0.6z⁻¹ + 0.1z⁻²). Design a digital high pass filter with ω'c = 0.3567π. ## Short answer Apply the transformation z⁻¹ → −(z⁻¹+α)/(1+αz⁻¹) with α=−0.5766. Result: H'(z)=(0.4811−0.9433z⁻¹+0.3839z⁻²)/(1−0.6824z⁻¹+0.1259z⁻²). ## Answer The digital low-pass to high-pass transformation replaces z⁻¹ by z⁻¹ → −(z⁻¹ + α) / (1 + α z⁻¹), α = −cos[(ωc + ω'c)/2] / cos[(ωc − ω'c)/2] with ωc = 0.2575π (given low-pass) and ω'c = 0.3567π (required high-pass). #### Step 1: α (ωc + ω'c)/2 = 0.3071π and (ωc − ω'c)/2 = −0.0496π. α = −cos(0.3071π) / cos(0.0496π) = −0.5696 / 0.9879 = **-0.5766** #### Step 2: substitute H(z) = (0.1 + 0.4z⁻¹)/(1 − 0.6z⁻¹ + 0.1z⁻²). Put z⁻¹ = −(z⁻¹ + α)/(1 + αz⁻¹) and multiply numerator and denominator by (1 + αz⁻¹)²: Numerator: 0.1(1 + αz⁻¹)² − 0.4(z⁻¹ + α)(1 + αz⁻¹) Denominator: (1 + αz⁻¹)² + 0.6(z⁻¹ + α)(1 + αz⁻¹) + 0.1(z⁻¹ + α)² With α = -0.5766: ```text Numerator = 0.3306 − 0.6483 z⁻¹ + 0.2639 z⁻² Denominator = 0.6873 − 0.4690 z⁻¹ + 0.0865 z⁻² ``` Dividing by the constant term of the denominator (0.6873): **H'(z) = (0.4811 − 0.9433z⁻¹ + 0.3839z⁻²) / (1 − 0.6824z⁻¹ + 0.1259z⁻²)** #### Check - Low-pass H: |H(0)| = 1.000, |H(0.2575π)| = 0.7079, |H(π)| = 0.176. - High-pass H': |H'(0)| = 0.176, |H'(0.3567π)| = 0.7079 (the same gain as the low-pass at its cut-off 0.2575π), |H'(π)| = 1.000. So the new filter is high-pass with the same gain at its edge as the low-pass has at its edge. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which replacement gives low-pass to high-pass in digital domain? A. z^-1 -> (z^-1 + alpha)/(1 + alpha z^-1) B. z^-1 -> -(z^-1 + alpha)/(1 + alpha z^-1) C. z^-1 -> -z^-1 D. s -> 1/s Answer: B. The minus sign flips the response from low to high. (Key point: Substitution) 2. How is alpha found? A. alpha = cos[(wc + wc')/2]/cos[(wc - wc')/2] B. alpha = sin(wc)/sin(wc') C. alpha = -cos[(wc + wc')/2]/cos[(wc - wc')/2] D. alpha = wc/wc' Answer: C. This ensures the old edge maps to the new edge. (Key point: Alpha formula) 3. After substituting, what is multiplied through to clear fractions? A. (1 - alpha z^-1) B. z^2 C. (z^-1 + alpha) D. (1 + alpha z^-1)^2 Answer: D. The filter is second order, so the denominators are squared. (Key point: Algebra step) 4. What indicates H' is high-pass? A. Small gain at 0 and gain 1 at pi, with edge gain matching the low-pass edge B. Gain 1 at 0 C. Gain 0 at pi D. Gain equal everywhere Answer: A. The roles of 0 and pi swap. 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